It appears simple but I am concerned about the rates that each function approaches $\displaystyle \infty$. Should I be?

Given:

$\displaystyle 1+R=1+(T-t)L(t,T)$

and

$\displaystyle 1+R=e^{(T-t)y(t,T)$

show that

$\displaystyle \lim_{T \rightarrow t}L(t,T) = \lim_{T \rightarrow t}y(t,T)$.

PROOF:

Rearrange both formulae to get:

$\displaystyle L(t,T)=\frac{R}{(T-t)}$

and

$\displaystyle y(t,T)=\frac{ln(1+R)}{(T-t)}$

Take the limit of each as [tex] T \rightarrow t:

$\displaystyle \lim_{T \rightarrow t}L(t,T)=\lim_{T \rightarrow t}\frac{R}{(T-t)} = \infty$

and

$\displaystyle \lim_{T \rightarrow t}y(t,T)=\lim_{T \rightarrow t}\frac{ln(1+R)}{(T-t)} = \infty$

As both approach $\displaystyle \infty$ we can state that:

$\displaystyle \lim_{T \rightarrow t}L(t,T) = \lim_{T \rightarrow t}y(t,T)$.